Differental Equation Question, Particular Solutions?

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Discussion Overview

The discussion revolves around solving a differential equation of the form dy/dx = x^2 + x^2y with the initial condition y(1) = 2. Participants explore the steps to find the particular solution and clarify the use of the initial condition in determining the constant.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant expresses uncertainty about how to apply the initial condition y(1) = 2 after solving the differential equation.
  • Another participant confirms the initial steps taken and suggests using a constant A instead of C in the general solution.
  • A later reply proposes a method to solve for the constant A using the initial condition, but there is confusion about the correct equation to use.
  • Participants discuss the implications of plugging in values for x and y to find the constant, with some suggesting that the approach taken may have errors.
  • There is a suggestion to replace e^{x^3/3} with a numerical approximation, while others question the necessity of this step.
  • One participant reflects on the process of determining the specific value of y based on the initial condition, indicating a lack of clarity on how to proceed.

Areas of Agreement / Disagreement

Participants generally agree on the need to use the initial condition to find the constant but disagree on the specifics of how to apply it correctly. Confusion remains about the correct formulation of the equation and the final expression for y.

Contextual Notes

There are unresolved questions regarding the correct application of the initial condition and the formulation of the solution, as well as potential misinterpretations of the steps involved in solving for the constant.

stripedcat
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[math]dy/dx = x^2+x^2y, y(1)=2[/math]

I'm not sure what to do with the y(1)=2, presumably I want to do the equation and then find y and c?

[math]dy/dx=x^2(y+1)[/math]

[math]dy/y+1=x^2 dx[/math]

[math] ln(y+1)=x^3/3+C[/math]

[math]y=e^{x^3/3} - 1 + C[/math]

I'm assuming I did this part right?

I just plug in 2 where the y is and I where the x is, then solve for y and c? Is that how it works?
 
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Re: Differiental Equation Question, Particular Solutions?

stripedcat said:
[math]dy/dx = x^2+x^2y, y(1)=2[/math]

I'm not sure what to do with the y(1)=2, presumably I want to do the equation and then find y and c?

[math]dy/dx=x^2(y+1)[/math]

[math]dy/y+1=x^2 dx[/math]

[math] ln(y+1)=x^3/3+C[/math]

Correct so far.
[math]y=e^{x^3/3} - 1 + C[/math]
This should be $y=Ae^{x^3/3}-1$ where $A$ is some constant.

Now just use the initial condition to determine $A$.
 
Re: Differiental Equation Question, Particular Solutions?

Pranav said:
Correct so far.

This should be $y=Ae^{x^3/3}-1$ where $A$ is some constant.

Now just use the initial condition to determine $A$.

So like this...?

[math]2=Ce^{1^3/3}-1[/math]

Solve for C

[math]e^{1/3} = 1.3956[/math]

[math]2=C(1.3956)-1[/math]

[math]3=C(1.39561)[/math]

C=~2.1496

Now that I have C...

[math]y=(2.1496)(1.3956)-1[/math]

y=-0.24601?
 
Last edited:
Re: Differiental Equation Question, Particular Solutions?

stripedcat said:
So like this...?

[math]2=Ce^{1^3/3}[/math]

Shouldn't that be $3=Ce^{1/3}$?
 
Re: Differiental Equation Question, Particular Solutions?

Pranav said:
Shouldn't that be $3=Ce^{1/3}$?

Forgot the -1

But beyond that did I plug in the values like I was suppose to?
 
Re: Differiental Equation Question, Particular Solutions?

stripedcat said:
Now that I have C...

[math]y=(2.1496)(1.3956)-1[/math]

y=-0.24601?

Any reason to replace $e^{x^3/3}$ with $1.3956$? You got $C$ in a correct way, so the final answer is $y=1.3956e^{x^3/3}-1$, do you see why?
 
Re: Differiental Equation Question, Particular Solutions?

So I got C, that's good then! Thank you!

Do we not plug in the previous X value when trying to solve for Y? I was wondering about that, it didn't seem right, but then by not including it I didn't see how to solve for Y without it being an abstract answer rather than a specific one.

So it's just the [math]y=e^{x^3/3} -1[/math] for y then. Okay.
 
Re: Differiental Equation Question, Particular Solutions?

stripedcat said:
Do we not plug in the previous X value when trying to solve for Y? I was wondering about that, it didn't seem right, but then by not including it I didn't see how to solve for Y without it being an abstract answer rather than a specific one.
I am not sure what you ask here. The condition $y(1)=2$ was given to determine $A$. $y(1)=2$ means that at $x=1$, $y=2$, so you plug in these values of $x$ and $y$ and find $A$.
So it's just the [math]y=e^{x^3/3} -1[/math] for y then. Okay.

No, look at my previous post.
 

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